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mojhsa [17]
3 years ago
14

Use substitution to solve the linear system of equations.

Mathematics
1 answer:
USPshnik [31]3 years ago
8 0

Answer:

(- 1, 5 )

Step-by-step explanation:

substitute y = 5 into the other equation

6x + 20 = 14 ( subtract 20 from both sides )

6x = - 6 ( divide both sides by 6 )

x = - 1 and y = 5 ← given

solution (- 1, 5 )


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2x^2=9-3x <br> Find the factors
d1i1m1o1n [39]

Answer:

x = -3, $ \frac{3}{2} $

Step-by-step explanation:

The given quadratic equation is: $ 2x^2 = 9 - 3x $

This can be written as: $ 2x^2 + 3x - 9 = 0 $

To solve a quadratic equation of the form $ ax^2 + bx + c = 0 $ we use the formula:

           $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

Here, a = 2; b = 3; c = - 9

Therefore, the roots of the equation are:

$ x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)} $

$ \implies x = \frac{-3 \pm \sqrt{81}}{4} $

$ \implies x = \frac{-3 \pm 9}{4} $

We get two values of 'x', viz.,

x = $ \frac{-3 + 9}{4} $ and $ \frac{- 3 - 9}{4} $

$ \implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4} $

⇒ x = -3, 3/2

Since the factors of the quadratic equation is asked, we write it as:

(x + 3)(x - $ \frac{3}{2} $) = 0

because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.

Multiply (x + 3) and (x - $ \frac{3}{2} $ to see that this indeed is the given quadratic equation.

5 0
3 years ago
7th grade help me plzzzzz
Gekata [30.6K]

Answer:

fifteen, "number sentence"

Step-by-step explanation:

5 + 5 +5 = 15

4 0
3 years ago
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Step-by-step explanation:

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Step-by-step explanation:

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